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प्रश्न
Integrate the following with respect to x.
`1/(9 - 8x - x^2)`
उत्तर
`int 1/(9 - 8x - x^2) "d"x`
Consider `9 - 8x - x^2 = 3x^2 - (x^2 + 8x)`
= `9 - [(x + 4)^2 - 16]`
= `9 + 16 - (x + 4)^2`
= `25 - (x + 4)^2`
= `5^2 - (x + 4)^2`
So integral becomes
`int ("d"x)/(5^2 - (x + 4)^2) = 1/10 log|(5 + x + 4)/(5 - x - 4)| + "c"`
= `1/10 log|(9 + x)/(1 - x)| + "c"`
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